Question

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

Answer 1

\[\text{Let I }=\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

\[= \int_\frac{1}{3}^1 \frac{\left[ x^3 \left( \frac{x}{x^3} - 1 \right) \right]^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{x \left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{\left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^3}dx\]

Put

\[\left( \frac{1}{x^2} - 1 \right) = z\]

\[\therefore - \frac{2}{x^3}dx = dz\]
\[ \Rightarrow \frac{dx}{x^3} = - \frac{dz}{2}\]

When

\[x \to \frac{1}{3}, z \to 8\]

When

\[x \to 1, z \to 0\]

\[\therefore I = - \frac{1}{2} \int_8^0 z^\frac{1}{3} dz\]
\[ = \left.- \frac{1}{2} \times \frac{z^\frac{4}{3}}{\frac{4}{3}}\right|_8^0 \]
\[ = - \frac{3}{8}\left[ 0 - \left( 8 \right)^\frac{4}{3} \right]\]
\[ = - \frac{3}{8} \times \left( - 16 \right)\]
\[ = 6\]